Area of rhombus and interior isosceles triangles Points $E$, $F$, $G$, and $H$ lie inside a rhombus $ABCD$, such that the triangles $\triangle AEB$, $\triangle BHC$, $\triangle CGD$, and $\triangle DFA$ are isosceles right triangles with hypotenuses $AB$, $BC$, $CD$, and $DA$. The sum of areas of $ABCD$ and $EFGH$ is $S$. Find, with the proof, the length of $CD$. Express your answer in terms of $S$ only.
I know the area of the rhombus $\left( \frac12 d_1d_2 \right)$ and I know the four triangles are equivalent and can find their areas to subtract from $ABCD$. I'm unsure how to go about finding the remaining area (four triangles) to get the area of $EFGH$. My approach could be wrong though.
 A: WARNING: NOT RIGOROUS (intended to walk someone through how a proof would be constructed in an environment more conducive to diagrams). 
First, we note that if $ABCD$ is a square, then $EFGH$ collapses to a single point at the center. Let $\theta$ be the the smaller angle of the rhombus $ABCD$ and $x=CD$ be its side length. We see that $EFGH$ is a square since $$|EF|=|FG|=|GH|=|HE|=2\frac{x}{\sqrt{2}}\sin\left(\frac{\theta-\frac{\pi}{2}} {2}\right)$$ and the perpendicular bisectors of $EF$ and $GH$ are themselves perpendicular. Hence $$EFGH=2x^2\sin^2\left(\frac{\theta-\frac{\pi}{2}}{2}\right)=x^2(1-\sin(\theta))$$ $ABCD$ can similarly be expressed in terms of $\theta$ and $x$. Let the longer side of the rhombus be $d_1$, the shorter $d_2$. Then $d_1=2x\cos\left(\frac{\theta}{2}\right)$ and $d_2=2x\sin\left(\frac{\theta}{2}\right)$. Therefore $$ABCD=\frac{1}{2}\left(4x^2\left[\cos\left(\frac{\theta}{2}\right)\sin\left(\frac{\theta}{2}\right)\right]\right)=x^2\sin(\theta)$$ and $S=x^2$.
A: This question is much easier to deal with if you have a picture.
Let $\theta_{XYZ} = |\angle XYZ|$. That is, $\theta_{XYZ}$ denotes the measured size of $\angle XYZ$.
Without loss of generality, choose point labeling so that $\theta_{ABC}\leq\theta_{BCD}$.

Let $L$ denote the length of $\overline{CD}$, noting that all the sides of rhombus $ABCD$ have the same length. Note that the area of $ABCD$ is $L^2\sin\theta_{ABC}$.
Since the $4$ interior right triangles are all identical, the $4$ interior isosceles triangles ($\triangle AEF$, $\triangle BEH$, $\triangle CGH$, and $\triangle DFG$) are identical as well. Since the diagonals of $ABCD$ are perpendicular bisectors of the bases of the isosceles triangles that they cross, it follows that $EFGH$ is a square. Each side of the square has length $L\sqrt2\sin\frac{\theta_{EBH}}2$, therefore its area is $L^22\sin^2\frac{\theta_{EBH}}2 = L^2(1-\cos\theta_{EBH})$. Since $\theta_{EBH} = \frac\pi2 - \theta_{ABC}$, the area can be rewritten as $L^2 - L^2\sin\theta_{ABC}$.
As $S$ is the sum of the areas of $ABCD$ and $EFGH$, then $S = L^2$, thus $L = \sqrt S$.

A hint to the final answer could be obtained by considering the two degenerative cases: $\theta_{ABC} = \frac\pi2$ and $\theta_{ABC} = \frac\pi4$ (we can discount $\theta_{ABC} < \frac\pi4$, because then the points of $EFGH$ would lie outside the rhombus, although the result remains valid so long as the relative positions of where the points of $EFGH$ go are specified differently than just "inside").
For $\theta_{ABC} = \frac\pi2$, it is clear $ABCD$ is a square, and $EFGH$ collapses to a point, so $L = \sqrt S$.
For $\theta_{ABC} = \frac\pi4$, it is easier to see that $EFGH$ is a square and how to compute the length of the sides, since the sides of the interior right triangles are colinear with the sides of the outer rhombus.

A rigorous explanation of why $EFGH$ must be a square is left as an exercise.
